# pointwise convergence does not imply uniform convergence for series

Regarding sequences of functions $$(f_n(x))$$, I can wrap my head around the idea that uniform convergence $$\Rightarrow$$ pointwise convergence, but pointwise convergence does not imply uniform convergence.

However, regarding series of functions $$\sum f_n(x)$$, I am not sure if I totally understand. I have proved that if a series converges uniformly on $$\mathbb R$$, then it is pointwise convergence on $$\mathbb R$$ and $$\sup |f_n(x)|$$ converges to 0 as n goes to infinity. But I am not sure of the converse. I was pretty sure the converse did not hold true, but I cannot find a counter example.

Is it similar to sequences and the converse does not hold true? Or am I thinking of it wrong and there is a way to prove that it does also hold true?

• $\sum ((-1)^{n}/n)|x|$ converges pointwise on $\mathbb R$ but not uniformly. May 17, 2021 at 8:46
• How did I not think of that! It's a great counterexample. Thanks! May 17, 2021 at 8:54
• It is not true that uniform convergence $\Rightarrow \limsup f_n(x) = 0$, or that $\limsup f_n(x) = 0\Rightarrow$ uniform convergence. May 17, 2021 at 9:04
• The sequence $(99,99,99,\ldots)$ converges uniformly but not to zero. (Having said that, it's not clear to me what $\limsup f_n(x) = 0$ even means; perhaps you could clarify that?) May 17, 2021 at 9:21
• There is really no difference between sequences and series, because you can convert between them by simply defining the sequence $(g_n(x))$ as $g_n(x)=\sum_{i=0}^nf_n(x)$. Then all those theorems about uniform convergence become equivalent. May 17, 2021 at 20:29

The series $$\sum_{n=0}^\infty x^n$$ converges pointwise to $$\frac1{1-x}$$ on $$(-1,1)$$. However, the convergence is not uniform, since each function $$\sum_{n=0}^N x^n$$ is bounded on $$(-1,1)$$, but $$x\mapsto\frac1{1-x}$$ isn't.

• Ah. That makes sense. So a counter example exists since $\sum x^n$ converges uniformly only on a bounded interval. Thank you! May 17, 2021 at 8:52
• @interestingmeatloaf: That's not right $-$ $(0,1)$ is bounded. (I suppose that means that strictly speaking, what you wrote is correct. For instance, $\sum x^n$ doesn't converge on the unbounded interval $(99,\infty)$. But it's not what you meant!) May 17, 2021 at 9:24